如何在Node.js中获取脚本的路径?

我知道有流程。Cwd,但它只引用调用脚本的目录,而不是脚本本身。例如,假设我在/home/kyle/目录下,然后运行以下命令:

node /home/kyle/some/dir/file.js

如果我调用process.cwd(),我会得到/home/kyle/,而不是/home/kyle/some/dir/。有办法得到那个目录吗?


当前回答

如果你使用pkg来打包你的应用,你会发现这个表达式很有用:

appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));

process.pkg tells if the app has been packaged by pkg. process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app. require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode. __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty. For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.

其他回答

var settings = 
    JSON.parse(
        require('fs').readFileSync(
            require('path').resolve(
                __dirname, 
                'settings.json'),
            'utf8'));

基本上你可以这样做:

fs.readFile(path.resolve(__dirname, 'settings.json'), 'UTF-8', callback);

使用resolve()而不是连接'/'或'\',否则您将遇到跨平台问题。

注意:__dirname是模块或包含脚本的本地路径。如果你正在编写一个插件,需要知道主脚本的路径,它是:

require.main.filename

或者,获取文件夹名称:

require('path').dirname(require.main.filename)

如果你使用pkg来打包你的应用,你会发现这个表达式很有用:

appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));

process.pkg tells if the app has been packaged by pkg. process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app. require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode. __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty. For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.

I know this is pretty old, and the original question I was responding to is marked as duplicate and directed here, but I ran into an issue trying to get jasmine-reporters to work and didn't like the idea that I had to downgrade in order for it to work. I found out that jasmine-reporters wasn't resolving the savePath correctly and was actually putting the reports folder output in jasmine-reporters directory instead of the root directory of where I ran gulp. In order to make this work correctly I ended up using process.env.INIT_CWD to get the initial Current Working Directory which should be the directory where you ran gulp. Hope this helps someone.

var reporters = require('jasmine-reporters');
var junitReporter = new reporters.JUnitXmlReporter({
  savePath: process.env.INIT_CWD + '/report/e2e/',
  consolidateAll: true,
  captureStdout: true
 });

每个Node.js程序在其环境中都有一些全局变量,这些变量表示关于进程的一些信息,其中一个是__dirname。